Show that U is orthogonal U superscript T U. Where U is [-1/√2,1/√2,0; -1/√2,1/√2,0;1/√3,1/√3,1/√3]

To show that a matrix \( U \) is orthogonal, we need to verify that \( U^T U = I \), where \( U^T \) is the transpose of \( U \) and \( I \) is the identity matrix. 

Given matrix \( U \):
\[
U = \begin{pmatrix}
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}
\end{pmatrix}
\]

### Step 1: Find \( U^T \)
The transpose \( U^T \) is obtained by flipping the matrix over its diagonal, swapping rows and columns:
\[
U^T = \begin{pmatrix}
\frac{-1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
0 & 0 & \frac{1}{\sqrt{3}}
\end{pmatrix}
\]

### Step 2: Compute \( U^T U \)
Now, we multiply \( U^T \) by \( U \):
\[
U^T U = \begin{pmatrix}
\frac{-1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
0 & 0 & \frac{1}{\sqrt{3}}
\end{pmatrix}
\begin{pmatrix}
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}
\end{pmatrix}
\]

Let's compute this multiplication element by element. The result should be a 3x3 matrix.

### Step 3: Multiply the matrices

1. First row, first column:
   \[
   \left( \frac{-1}{\sqrt{2}} \times \frac{-1}{\sqrt{2}} \right) + \left( \frac{-1}{\sqrt{2}} \times \frac{-1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{3} = 1
   \]
   
2. First row, second column:
   \[
   \left( \frac{-1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) + \left( \frac{-1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} \right) = -\frac{1}{2} - \frac{1}{2} + \frac{1}{3} = 0
   \]

3. First row, third column:
   \[
   \left( \frac{-1}{\sqrt{2}} \times 0 \right) + \left( \frac{-1}{\sqrt{2}} \times 0 \right) + \left( \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} \right) = 0
   \]

4. Second row, first column:
   \[
   \left( \frac{1}{\sqrt{2}} \times \frac{-1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{2}} \times \frac{-1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} \right) = 0
   \]

5. Second row, second column:
   \[
   \left( \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} \right) = 1
   \]

6. Second row, third column:
   \[
   0
   \]

7. Third row, first column:
   \[
   0
   \]

8. Third row, second column:
   \[
   0
   \]

9. Third row, third column:
   \[
   1
   \]

### Step 4: Final Result

The resulting product is:
\[
U^T U = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} = I
\]

Since \( U^T U = I \), this shows that \( U \) is an orthogonal matrix.

Would you like a deeper explanation of any specific step?

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**Relative questions:**
1. What are the properties of orthogonal matrices?
2. Can an orthogonal matrix have complex elements?
3. How do orthogonal matrices relate to rotation in vector spaces?
4. How do you compute the inverse of an orthogonal matrix?
5. What is the geometric interpretation of the matrix \( U \)?

**Tip:** Orthogonal matrices preserve the length of vectors and angles between them during transformation.

Learn how to demonstrate that matrix U is orthogonal using matrix algebra techniques. This problem involves verifying the property U^T U = I for matrix U, which is given as [-1/√2, 1/√2, 0; -1/√2, 1/√2, 0; 1/√3, 1/√3, 1/√3]. Suitable for advanced undergraduate students studying matrix algebra and linear algebra.

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